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If I do something like:

$$\frac{dy}{dx} = D$$

$$dy = D \times dx$$

People would often say that it is not rigorous to do so. But if we start from the definition of the derivative:

$$\lim_{h \to 0}{\frac{f(x + h) - f(x)}{h}} = D$$

And by using the properties of limits we can say:

$$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$

And then finally:

$$\lim_{h \to 0}(f(x + h) - f(x)) = D \times (\lim_{h \to 0} h)$$

Isn't this the same? Or am I missing something?

Anonymous
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    Your claim after "And by using the properties of limits we can say:" is false. – smcc Jul 15 '16 at 14:06
  • $dx$ and $dy$ are not variables. They're an operation like addition, subtraction, integration, differentiation, etc. –  Jul 15 '16 at 14:06
  • @KingDuken Even if I define dx and dy to be the denominator and numerator of the fraction (of the definition of the derivative)? – Anonymous Jul 15 '16 at 14:09
  • @smcc How so? I do not understand. – Anonymous Jul 15 '16 at 14:12
  • What do you [b]mean[/b] by "the upper and lower part of the fraction (of the definition of the derivative"? In the definition of the derivative, "dy/dx" is NOT a fraction. It is, rather, the limit of the fraction (f(x+ h)- f(x))/h. – user247327 Jul 15 '16 at 14:12
  • Why the downvote? – Anonymous Jul 15 '16 at 14:13
  • You haven't defined what $dy/dx$ was except that it's $D$, which is a normal notation for a derivative. Henceforth, $D$ is also an operation, not a variable. –  Jul 15 '16 at 14:14
  • You can "distribute" the limit over the quotient if and only if the limits in the numerator and denominator both exist and the limit in the denominator is not zero. But the limit in the denominator is zero, so your "distributed" form is 0/0, which isn't defined. – Ian Jul 15 '16 at 14:15
  • As for your very last equation, it is actually true (within the standard framework), but somewhat trivially: the left side is zero and the right side is also zero. Consequently that equation does not pin down what $D$ has to be. As it happens it doesn't even pin down the right criterion for $D$ to exist, because that equation holds under the assumption of just continuity. – Ian Jul 15 '16 at 14:16
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    Anyway, there actually are nonstandard systems out there that let you manipulate infinitesimals rigorously. The most common one is "hyperreal analysis". You can get a gentle introduction to this in a calculus book by Keisler. But it doesn't look very much like you might naively expect. – Ian Jul 15 '16 at 14:19
  • This isn't a full answer, but what is $\lim\limits_{h\to 0}h$? Of course it is $0$. Right off the bat, this should tell you something is wrong since you can't divide by $0$. – pancini Jul 15 '16 at 14:25
  • @Anonymous Forgetting about your clear misconceptions with limits, you can treat $dx$ and $dy$ as variables. One way to do it is to say that given a function $x\mapsto f(x)$, its differential is given by $df(x,h) = f'(x)dx(x,h)$ (or $df=f'(x)dx$ when we suppress the arguments of the differentials) , where $dx = (x+h)-x$ for some small $h\ne 0$. Then the map $x\mapsto x$ has the differential $dx = 1dx$ (obviously) and $x\mapsto y$ has the differential $dy=y'dx$. Thus $y' = \frac{dy}{dx} = \frac{y'dx}{dx}$. –  Jul 15 '16 at 14:35
  • The quick answer is this: in order for your last two equations to work, $\lim_{h \to 0}h$ needs to be an "infinitesimal (infinitely small) quantity, which is not equal to zero". Under the usual conception of real numbers, there is no such thing. There are, however, modern frameworks that allow infinitesimal quantities wherein these quantities can be divided to produce derivatives in that intuitive sense. – Ben Grossmann Jul 15 '16 at 14:52
  • @Bye_World Except now you've assigned a different meaning to the symbols than the one the OP expects. In particular, your $\frac{dy}{dx}$ is an approximate derivative, not a derivative. – Ian Jul 15 '16 at 14:56
  • @Ian In what way is it approximate? $h\mapsto f'(x_0)h$ is just the derivative seen as a linear map rather than a number. That's essentially what I defined above. Then algebraically $\frac{f'(x_0)h}{h} = f'(x_0)$. The only little caveat is that $h\ne 0$. –  Jul 15 '16 at 15:03
  • Er, sorry, I missed the $f'(x)$ in your original definition of $df$. Now the subtlety is that your $dy$ is just being defined as $f' dx$ where $f'$ is defined the usual way and $dx$ is just $h$. Thus you haven't done anything to understand the derivative itself in this framework, you've just written down a linear map associated to the derivative. So it isn't really "using differentials as variables", at least not the way the OP appears to mean. – Ian Jul 15 '16 at 15:05
  • (Cont.) This is part of why I've never liked this little line of discourse in calculus books (even though I acknowledge that it is eventually useful in differential geometry). – Ian Jul 15 '16 at 15:06
  • @Omnomnomnom Can you specify what "modern frameworks" are you talking about? I want to read about it. – Anonymous Jul 15 '16 at 15:09
  • @Anonymous here's the wiki page for "non-standard analysis", which is the most prominent example. Be sure to look at the question that I linked to. – Ben Grossmann Jul 15 '16 at 15:15

1 Answers1

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I can spot the following mistake in your attempt:

  1. And by using the properties of limits we can say:

    $$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$

You cannot actually do that, as smcc has said. You must note that $$\lim_{x\to 0} \frac{f(x)}{g(x)}=\frac{\lim_\limits{x\to 0} f(x)}{\lim_\limits{x\to 0} g(x)} \,\,\,\,\,\,\,\,\,\,\mathrm {iff \lim_\limits{x\to 0} g(x) \not = 0}$$ So what you have said is not right.

Now coming to the actual question, if $dx$ and $dy$ can be treated as variables, most of the mathematicians treat $\frac{d}{dx}$ as a mathematical operator (like $+,-,*,/$) which acts on the variable $y$. So that way, you can clearly understand what is the variable and what is not.

However, if you are strict enough to observe from the "limit" viewpoint, then observe that $\frac{dy}{dx}$ is nothing but $\lim_\limits{\Delta x\to 0}\frac{\Delta y}{\Delta x}$. Now $\frac{\Delta y}{\Delta x}$ is a fraction with $\Delta y$ and $\Delta x$ in the numerator and denominator. So you can view them as variables now.

Looks a bit weird, I know, but it entirely depends on how you want to support your argument and from which point of view you want to make your claim.

SchrodingersCat
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    $D$ is not for the euler's notation. It is just a number. The value of the derivative. – Anonymous Jul 15 '16 at 15:00
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    Using $D$ as the letter for the numerical value of the derivative is a bit contrary to certain standard notation (e.g. the notation used in Evans), but it is clear in this context. – Ian Jul 15 '16 at 15:01
  • @Anonymous Okay, I have got it. Going to edit my answer. But tell me, is the derivative equal to D for all $x$ or just some $x=x_0$? – SchrodingersCat Jul 15 '16 at 15:05
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    @SchrodingersCat for all x's. A general one. – Anonymous Jul 15 '16 at 15:08